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Example 30 - Find coordinates of point where line through A (3, 4, 1)

Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Example 30 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

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Example 30 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points with position vectors π‘Ž βƒ— & 𝑏 βƒ— is 𝒓 βƒ— = 𝒂 βƒ— + πœ† (𝒃 βƒ— βˆ’ 𝒂 βƒ—) Given the line passes through the points (𝑏 βƒ— βˆ’ π‘Ž βƒ—) = (5𝑖 Μ‚ + 1𝑗 Μ‚ + 6π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚) = 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚ A (3, 4, 1) π‘Ž βƒ— = 3𝑖 Μ‚ + 4𝑗 Μ‚ + π‘˜ Μ‚ B (5, 1, 6) 𝑏 βƒ— = 5𝑖 Μ‚ + 1𝑗 Μ‚ + 6π‘˜ Μ‚ ∴ π‘Ÿ βƒ— = (3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚) + πœ† (2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 5π‘˜ Μ‚) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, π‘Ÿ βƒ— = x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ = 3𝑖 Μ‚ + 4𝑗 Μ‚ + 1π‘˜ Μ‚ + 2πœ†π‘– Μ‚ βˆ’ 3πœ†π‘— Μ‚ + 5πœ†π‘˜ Μ‚ x𝑖 Μ‚ + y𝑗 Μ‚ + 0π‘˜ Μ‚ = (3 + 2πœ†)𝑖 Μ‚ + (4 βˆ’ 3πœ†)𝑗 Μ‚ + (1 + 5πœ†)π‘˜ Μ‚ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5πœ† ∴ πœ† = (βˆ’πŸ)/πŸ“ So, x = 3 + πœ† = 3 + 2 Γ— (βˆ’1)/5 = 3 βˆ’ 2/5 = 13/5 & y = 4 βˆ’ 3πœ† = 4 βˆ’ 3 Γ— (βˆ’1)/5 = 4 + 3/5 = 23/5 Therefore, the required coordinates are (πŸπŸ‘/πŸ“,πŸπŸ‘/πŸ“,𝟎) Example 30 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane.The equation of a line passing through two points A(π‘₯_1, 𝑦_1, 𝑧_1) and B(π‘₯_2, 𝑦_2, 𝑧_2) is (𝒙 βˆ’ 𝒙_𝟏)/(𝒙_𝟐 βˆ’ 𝒙_𝟏 ) = (π’š βˆ’ π’š_𝟏)/(π’š_𝟐 βˆ’ π’š_𝟏 ) = (𝒛 βˆ’ 𝒛_𝟏)/(𝒛_𝟐 βˆ’ 𝒛_𝟏 ) Given the line passes through the points So, the equation of line is (π‘₯ βˆ’ 3)/(5 βˆ’ 3) = (𝑦 βˆ’ 4)/(1 βˆ’ 4) = (𝑧 βˆ’ 1)/(6 βˆ’ 1) A (3, 4, 1) ∴ π‘₯_1= 3, 𝑦_1= 4, 𝑧_1= 1 B(5, 1, 6) ∴ π‘₯_2 = 5, 𝑦_2= 1, 𝑧_2= 6 (π‘₯ βˆ’ 3)/2 = (𝑦 βˆ’ 4)/(βˆ’3) = (𝑧 βˆ’ 1)/5 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = βˆ’1 ∴ k = (βˆ’πŸ)/πŸ“ So, x = 2k + 3 = 2 Γ— (βˆ’1)/5 + 3 = 3 βˆ’ 2/5 = 13/5 y = βˆ’3 Γ— (βˆ’1)/5 + 4 = 4 + 3/5 = 23/5 Therefore, the coordinates of the required point are (πŸπŸ‘/πŸ“, πŸπŸ‘/πŸ“, 𝟎).

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.